3.5.90 \(\int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx\) [490]

3.5.90.1 Optimal result

Integrand size = 21, antiderivative size = 118 \[ \int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx=\frac {2 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}} \]

output

2*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+ 
1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos( 
d*x+c))^(1/2)+2*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic 
Pi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^ 
(1/2)/d/(a+b*cos(d*x+c))^(1/2)
 

3.5.90.2 Mathematica [A] (verified)

Time = 13.94 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+a \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{d \sqrt {a+b \cos (c+d x)}} \]

input

Integrate[Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x],x]
 

output

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b*EllipticF[(c + d*x)/2, (2*b)/(a + 
 b)] + a*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)]))/(d*Sqrt[a + b*Cos[c + 
 d*x]])
 

3.5.90.3 Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3282, 3042, 3142, 3042, 3140, 3286, 3042, 3284}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec (c+d x) \sqrt {a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3282

\(\displaystyle b \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx+a \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle b \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3142

\(\displaystyle a \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3140

\(\displaystyle a \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}\)

input

Int[Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x],x]
 

output

(2*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + 
b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + (2*a*Sqrt[(a + b*Cos[c + d*x])/(a + b) 
]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]])
 

3.5.90.3.1 Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3282

Int[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/((a_.) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[d/b   Int[1/Sqrt[c + d*Sin[e + f*x]], x], x 
] + Simp[(b*c - a*d)/b   Int[1/((a + b*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x 
]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 
 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3286

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]
 

3.5.90.4 Maple [A] (verified)

Time = 3.47 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.64

method	result	size
default	\(-\frac {2 \sqrt {\left (2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b -\Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right ) a \right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}\, d}\)	\(194\)

input

int(sec(d*x+c)*(a+cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)
 

output

-2*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*(EllipticF( 
cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b-EllipticPi(cos(1/2*d*x+1/2*c),2,( 
-2*b/(a-b))^(1/2))*a)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^ 
2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d
 

3.5.90.5 Fricas [F(-1)]

Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx=\text {Timed out} \]

input

integrate(sec(d*x+c)*(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")
 

output

Timed out
 

3.5.90.6 Sympy [F]

\[ \int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx=\int \sqrt {a + b \cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \]

input

integrate(sec(d*x+c)*(a+b*cos(d*x+c))**(1/2),x)
 

output

Integral(sqrt(a + b*cos(c + d*x))*sec(c + d*x), x)
 

3.5.90.7 Maxima [F]

\[ \int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]

input

integrate(sec(d*x+c)*(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")
 

output

integrate(sqrt(b*cos(d*x + c) + a)*sec(d*x + c), x)
 

3.5.90.8 Giac [F]

\[ \int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right ) \,d x } \]

input

integrate(sec(d*x+c)*(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")
 

output

integrate(sqrt(b*cos(d*x + c) + a)*sec(d*x + c), x)
 

3.5.90.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \cos (c+d x)} \sec (c+d x) \, dx=\int \frac {\sqrt {a+b\,\cos \left (c+d\,x\right )}}{\cos \left (c+d\,x\right )} \,d x \]

input

int((a + b*cos(c + d*x))^(1/2)/cos(c + d*x),x)
 

output

int((a + b*cos(c + d*x))^(1/2)/cos(c + d*x), x)